Exercise A0.V.2.19

Definition (discrete valuation)

A discrete valuation on a field $k$ is a surjective homomorphism of abelian groups $v : (k^{\times},\cdot) \to (\mathbb{Z},+)$ such that $v(a + b) \geq \min(v(a), v(b))$ for all $a,b \in k^{\times}$ such that $a + b \in k^{\times}.$

Proposition A

Let $k$ be a field and let $v$ be a discrete valuation on $k.$ Then the set $R = \{a \in k^{\times} \mid v(a) \geq 0\} \cup \{0\}$ is a subring of $k$ and, in fact, $R$ is a Euclidean domain.

Proof

Since group homomorphisms preserve identity, $v(1) = 0.$ Thus $1 \in R$ and we only need to check closure.

Suppose $a,b \in R$ are non-zero, then $v(ab) = v(a) + v(b)$ and a sum of two non-negative integers is itself non-negative. So $ab \in R.$ If, instead, either $a = 0$ or $b = 0,$ then $ab = 0,$ which is also in $R.$

Suppose $a, b \in R$ and $a + b = 0.$ In that case, $a + b \in R.$ Suppose otherwise, that $a + b \in k^{\times}.$ If $a = 0$ or $b = 0,$ the sum is equal to the other summand, which is in $R.$ If neither is $0,$ then $v(a + b) \geq \min(v(a),v(b))$ and since $v(a) \geq 0$ and $v(b) \geq 0,$ $v(a + b) \geq 0.$

Observe that $(-1) \cdot (-1) = 1$ making $-1$ an involution in $k^{\times}.$ Since $(\mathbb{Z},+)$ is torsion-free, $v(-1) = 0.$ Hence, for each $a \in R,$ the element $-a = -1 \cdot a$ has valuation of $v(-1) + v(a) = v(a),$ so $-a \in R.$

This demonstrates that $R$ is a subring of $k.$

Suppose $a, b \in R$ are non-zero. We want to find $q, r \in R$ such that $a = qb + r$ with $r = 0$ or $v(r) < v(b).$ Let $S = \{ (q',r') \in R \times R \mid a = q'b + r' \}.$ Note that $(0,a) \in S,$ so $S$ is non-empty. If $(q',0) \in S$ then we are done. Assume that all elements of $S$ carry a non-zero remainder.

Let $(q, r) \in S.$ Suppose, for the sake of contradiction, that $v(r) \geq v(b).$ Then $v(r) - v(b) \geq 0$ and hence $r/b \in R.$ In that case, $a = (q + r/b)b + 0$ which gives rise to $(q + r/b, 0) \in S,$ contradicting our assumption that all elements of $S$ have non-zero remainder.

Thus $v(r) < v(b)$ and hence $v$ is a Euclidean valuation.

$\blacksquare$

Proposition B

The ring of rational numbers with the denominator not divisible by a fixed prime $p$ is a DVR.

Proof

Let $v : \mathbb{Q}^{\times} \to (\mathbb{Z},+)$ be a homomorphism of abelian groups defined as $p \mapsto 1, q \mapsto 0, -1 \mapsto 0,$ where $q$ is a prime distinct from $p.$ Since $\mathbb{Q}^{\times} \cong (\mathbb{Z}/2\mathbb{Z}) \oplus \bigoplus_{p \in \mathbb{P}} \mathbb{Z},$ the universal property of a direct sum guarantees that $v$ is a well-defined surjective homomorphism.

Let $a, b \in \mathbb{Q}^{\times}$ such that $a + b \in \mathbb{Q}^{\times}.$ Write $a = p^m a'$ and $b = p^n b'$ such that $v(a') = v(b') = 0.$ Without loss of generality, assume that $n \leq m,$ hence $a + b = p^n (p^{m - n}a' + b').$ So $v(a + b) = \min(v(a), v(b)) + v(p^{m - n}a' + b').$

Let $\operatorname{den}(x)$ denote the reduced denominator of $x,$ and let $\mathsf{F}(x)$ denote the multiset of factors of $x$ (up to a consistent choice of representatives) as in 2.1.

Recall that the sum of two fractions $x/x' + y/y'$ can be written in reduced form as follows: start from $(xy' + yx')/(x'y')$ and cancel out the redundant factors. Thus,

$$\mathsf{F}(\operatorname{den}(x/x' + y/y')) \subseteq \mathsf{F}(x'y') = \mathsf{F}(x') + \mathsf{F}(y'),$$

that is, cancelling factors cannot add new factors into the denominator above what is already present in $\mathsf{F}(x'y').$

Since $m - n \geq 0,$ $p \notin \mathsf{F}(\operatorname{den}(p^{m - n}a')).$ Since $v(b') = 0,$ $p \notin \mathsf{F}(\operatorname{den}(b'))$ also. Certainly $p \notin \mathsf{F}(\operatorname{den}(p^{m - n}a')) + \mathsf{F}(\operatorname{den}(b')).$

Thus $v(p^{m - n}a' + b') \geq 0,$ and hence $v(a + b) \geq \min(v(a), v(b)).$

This gives us a discrete valuation on $\mathbb{Q}.$ Taking $0$ and all elements of $x \in \mathbb{Q}$ such that $v(x) \geq 0$ gives us a DVR. Since $v(x) \geq 0,$ $x$ (as a reduced fraction) has no powers of $p$ in the denominator.

$\blacksquare$